PLEASE HELP! :(

[hakke]

A hiker walks from point A on a bearing of 010◦ for 5 km and then on a bearing of 075◦ for 7 km to reach point B.
a   Find the length of AB.
b   Find the bearing of B from the start point A.
A second hiker travels from point A on a bearing of 080◦ for 4 km to a point P, and then travels in a straight line to B.
c Find:
i   the total distance travelled by the second hiker
ii   the bearing on which the hiker must travel in order to reach B from P.

A third hiker also travels from point A on a bearing of 080◦ and continues on that bearing until he reaches point C. He then turns and walks towards B. In doing so, the two legs of the journey are of equal length.
d   Find the distance travelled by the third hiker to reach B.

Please help with part b onwards? For part b), I used the sine rule and got an answer of ~51 degrees, and when I used the cos rule, I got an answer of 48.3 degrees. The answer is 49 degrees! :|

Could you please help with part c as well?

Thanks!

[hakke]

Edit: Really need help with part (ii) of (c) and also part (d)

Many thanks!

[brightsky]

Construct the triangle and call the point where he turns (between A and B) be C. So we have triangle ABC. Angle ACB = 115 (by simply putting in some given angles and working out unknowns). AC = 5, BC = 7. So by the cosine rule, AB^2 = 5^2 + 7^2 - 2*5*7 cos(115). So AB = 10.177587...By the sine rule, sin(115)/10.177587 = sin(angle CAB)/7, sin(angle CAB) = 7sin(115)/10.177587 = 0.62334..so angle CAB = arcsin(0.62334...) = 38.56 degrees, so the angle we want is 38.56 + 10 = 48.56 degrees = 49 degrees.

[hakke]

What about part (ii) of (c) and part (d)?

Thank you

[brightsky]

What about part (ii) of (c) and part (d)?

Thank you

Angle BAP = 70 - 38.56 = 31.44. We know AB = 10.177587... so by the cosine rule, BP^2 = 4^2 + 10.177587^2 - 2*4*10.177587*cos(31.44) so BP = 7.07928. So total distance is 11.07928. By the sine rule, sin(angle APB)/10.177587 = sin(31.44)/7.07928 so angle ABC = arcsin(0.74989) = 180 - 48.58 = 131.42. So the bearing is 131.42 - 90 - 10 = 31.42.

So angle BAC still = 31.44. Hence BC^2 = AC^2 + 10.177587^2 - 2*AC*10.177587*cos(31.44) but since BC = AC, then BC^2 = BC^2 + 10.177587^2 - 2*BC*10.177587*cos(31.44) ---> 10.177587^2 - 2*BC*10.177587*cos(31.44) --> x = 5.96445 so 2x = 11.8289.

[hakke]

angle ABC = arcsin(0.74989) = 180 - 48.58 = 131.42. So the bearing is 131.42 - 90 - 10 = 31.42.

I don't get this part?
angle APB = 48.58 degrees, so how does angle ABC = 180 - 48.58?

And then where did you get the -90 and -10 degrees from?

Thank you!

[brightsky]

We have sin(angle APB) = 10.177587*sin(31.44)/7.07928, where 90 < angle APB < 180. Since sin(pi - x) = sin(x), we want sin(pi - x).
For the -90 and -10, if you draw the diagram, you'll figure it out. It's hard to explain in words.

[hakke]

Why is angle APB in the 2nd quadrant?

Since angle APB works out to be 47.975 degrees? :S